# Squeeze Theorem/Functions/Proof 2

## Theorem

Let $a$ be a point on an open real interval $I$.

Let $f$, $g$ and $h$ be real functions defined at all points of $I$ except for possibly at point $a$.

Suppose that:

$\forall x \ne a \in I: g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$
$\displaystyle \lim_{x \mathop \to a} \ g \left({x}\right) = \lim_{x \mathop \to a} \ h \left({x}\right) = L$

Then:

$\displaystyle \lim_{x \mathop \to a} \ f \left({x}\right) = L$

## Proof

Let $f, g, h$ be real functions defined on an open interval $\left({a \,.\,.\, b}\right)$, except possibly at the point $c \in \left({a \,.\,.\, b}\right)$.

Let:

$\displaystyle \lim_{x \mathop \to c} \ g \left({x}\right) = L$
$\displaystyle \lim_{x \mathop \to c} \ h \left({x}\right) = L$
$g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$ except perhaps at $x = c$.

Let $\left \langle {x_n} \right \rangle$ be a sequence of points of $\left({a \,.\,.\, b}\right)$ such that:

$\forall n \in \N_{>0}: x_n \ne c$

and:

$\displaystyle \lim_{n \mathop \to \infty} \ x_n = c$
$\displaystyle \lim_{n \mathop \to \infty} \ g \left({x_n}\right) = L$

and:

$\displaystyle \lim_{n \mathop \to \infty} \ h \left({x_n}\right) = L$

Since:

$g \left({x_n}\right) \le f \left({x_n}\right) \le h \left({x_n}\right)$

it follows from the Squeeze Theorem for Real Sequences that:

$\displaystyle \lim_{n \mathop \to \infty} \ f \left({x_n}\right) = L$

The result follows from Limit of Function by Convergent Sequences.

$\blacksquare$